sin[sin-1() + cos-1(x)] = 1

Let a = sin-1() and b = cos-1(x)

Then sin(a) =  and cos(b) = x

And now our equation is 

sin(a + b) = 1

Using an identity on the left side,

sin(a)cos(b) + cos(a)sin(b) = 1

Draw two right triangles, one with acute angle a and one with acute angle b:



Since the sine of a is  and since , put
the numerator of , which is 1, on the side opposite a and 
the denominator of  which is 5, on the hypotenuse.

Since the cosine of b is x, we consider that as the fraction 
and since , put the numerator of , which is x,
on the side adjacent b and the denominator of  which is 1, on the
hypotenuse.




Next we use the Pythagorean theorem to find the side adjacent a, and
the side opposite b:

a² + b² = c²              a² + b² = c²
a² + 1² = 5²              x² + b² = 1²
 a² + 1 = 25              x² + b² = 1
     a² = 24                   b² = 1 - x² 
      a =                    b =              



Now we can use the right triangles above and their sides to substitute
into the equation:

sin(a)cos(b) + cos(a)sin(b) = 1

()() + ()() = 1

Clear of fractions by multiplying through by LCD 5

x +  = 5

Isolate the radical term:

     = 5 - x

Square both sides of the equations:

       24(1 - x²) = (5 - x)²

        24 - 24x² = (5 - x)(5 - x)

        24 - 24x² = 25 - 10x + x²

Get 0 on the right:

   25x² - 10x + 1 = 0

Factor the left side:

 (5x - 1)(5x - 1) = 0

        (5x - 1)² = 0 

           5x - 1 = 0

               5x = 1
                x =     

Edwin